Question

A survey of 400 non-fatal accidents showed that 189 involved the use of a cell phone. Find a point estimate for p, the population proportion of non-fatal accidents that involved the use of a cell phone. Find the 95% confidence interval for the proportion of non-fatal accidents that involved the use of a cell phone

Answer 1

Answer:

The point estimate for the proportion is p = 0.4725

The 95% confidence interval for the proportion of non-fatal accidents that involved the use of a cell phone is (0.4236, 0.5214).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 400$

Point estimate

$\pi = p = \frac{189}{400} = 0.4725$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4725 - 1.96\sqrt{\frac{0.4725*0.5275}{400}} = 0.4236$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4725 + 1.96\sqrt{\frac{0.4725*0.5275}{400}} = 0.5214$

The 95% confidence interval for the proportion of non-fatal accidents that involved the use of a cell phone is (0.4236, 0.5214).